About which axis of rotation is the radius of gyration of a body the least ?
Q 73
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The radius of gyration of a body is the least about an axis through the centre of mass (CM) of the body.
From the parallel axis theorem, we know that a given body has the smallest possible moment of inertia about an axis through its $\mathrm{CM}$. The radius of gyration of a body about a given axis is directly proportional to the square root of its moment of inertia about that axis. Hence, the conclusion.
$
\left\{O R I=I_{C M}+M h^2 . \therefore M k^2=\backslash\left([/ \text { latexM k_\{ } \operatorname{mathrm}\{C M\}\}^{\wedge}\{2\}\right]+M^2\right. \text {. }
$
$\therefore \mathrm{k}^2=\left[\right.$ latex] $\mathrm{k}_{-}\{\mathrm{mathrm}\{\mathrm{CM}\}\}^{\wedge}\{2\} \mathrm{N}+\mathrm{h}^2$, which shows that $\mathrm{k}$ is minimum, equal to $\mathrm{k}_{\mathrm{CM}}$ when $\mathrm{h}=0$.
From the parallel axis theorem, we know that a given body has the smallest possible moment of inertia about an axis through its $\mathrm{CM}$. The radius of gyration of a body about a given axis is directly proportional to the square root of its moment of inertia about that axis. Hence, the conclusion.
$
\left\{O R I=I_{C M}+M h^2 . \therefore M k^2=\backslash\left([/ \text { latexM k_\{ } \operatorname{mathrm}\{C M\}\}^{\wedge}\{2\}\right]+M^2\right. \text {. }
$
$\therefore \mathrm{k}^2=\left[\right.$ latex] $\mathrm{k}_{-}\{\mathrm{mathrm}\{\mathrm{CM}\}\}^{\wedge}\{2\} \mathrm{N}+\mathrm{h}^2$, which shows that $\mathrm{k}$ is minimum, equal to $\mathrm{k}_{\mathrm{CM}}$ when $\mathrm{h}=0$.
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